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3 months ago

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The surface is tilted to an angle of 37 degrees from the horizontal, as shown above in Figure 3. The blocks are each given a pus

h so that at the instant shown, they are moving toward each other. ii. At the instant shown in Figure 3, the blocks are moving toward each other with the same speed of 0.35m/s . The blocks collide 0.50 seconds later. What is the speed of the two-block system’s center of mass just before the blocks collide?

1 answer:

inna [3.1K]3 months ago

6 0

Answer:

Incomplete question: "Each block has a mass of 0.2 kg"

The velocity of the center of mass for the two-block system just prior to their collision is 2.9489 m/s

Explanation:

Provided information:

θ = angle of the surface = 37°

m = mass of each block = 0.2 kg

v = speed = 0.35 m/s

t = collision time = 0.5 s

Question: What is the velocity of the center of mass for the two-block system right before the blocks collide, vf =?

Change in momentum:

$delta(P)=F*delta(t)$

$P_{f} -P_{i}=F*delta(t)$

$2m(v_{f} -v_{i})=F*delta(t)$

$v_{i} =0.35-0.35=0$

It’s essential to calculate the required force:

$F=(m+m)*g*sin\theta$

Here, g = acceleration due to gravity = 9.8 m/s²

$F=(0.2+0.2)*9.8*sin37=2.3591N$

$v_{f} =\frac{F*delta(t)}{2m} =\frac{2.3591*0.5}{2*0.2} =2.9489m/s$

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